Epimorphisms of locally compact spaces

Solution 1:

Yes, a morphism in this category is epic iff it is surjective. Obviously surjective morphisms are epic, so it suffices to show that if $f: X\to Y$ is not surjective, there is a locally compact Hausdorff space $Z$ and distinct $g_1, g_2 \in \hom(Y,Z)$ such that $g_1 \circ f = g_2 \circ f$.

I will use the notion of a perfect map as defined in Engelking's General Topology, which is a continuous closed map from a Hausdorff space to a topological space with compact fibres. With this definition, perfect maps are proper (thm. 3.7.2) and compositions of perfect maps are perfect. (cor 3.7.3) In fact, a map from a Hausdorff space to a compactly generated Hausdorff space is perfect iff it is continuous and proper. (thm. 3.7.18)

On the product $Y \times 2$ define the equivalence relation $E$ such that $(x, i) \,E\, (y, j)$ iff $x = y$ and ($i = j$ or $x \in f(X)$). (In other words: glue the two copies of $f(X)$ together). Let $Z = (Y \times 2)/E$ and $q: Y \times 2 \to Z$ the natural map. Since the fibres of $q$ are finite they are compact and, using the fact that $f(X)$ is closed, it is easy to verify that if $F$ is closed in $Y \times 2$, so is $q^{-1}(q(F))$. Hence $q$ is perfect and since perfect images of locally compact Hausdorff spaces are also locally compact Hausdorff (thm. 3.7.20, 3.7.21), $q$ is a morphism.

If we now take $g_1$ and $g_2$ to be the compositions of $q$ with the two natural embeddings $Y \to Y \times 2$, we find they differ on $Y\setminus f(X)$, but agree on $f(X)$, therefore $ g_1 \circ f = g_2 \circ f$.